Difference between revisions of "2004 AMC 8 Problems/Problem 16"
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==Solution== | ==Solution== | ||
The first pitcher contains <math>600 \cdot \frac13 = 200</math> mL of orange juice. The second pitcher contains <math>600 \cdot \frac25 = 240</math> mL of orange juice. In the large pitcher, there is a total of <math>200+240=440</math> mL of orange juice and <math>600+600=1200</math> mL of fluids, giving a fraction of <math>\frac{440}{1200} = \boxed{\textbf{(C)}\ \frac{11}{30}}</math>. | The first pitcher contains <math>600 \cdot \frac13 = 200</math> mL of orange juice. The second pitcher contains <math>600 \cdot \frac25 = 240</math> mL of orange juice. In the large pitcher, there is a total of <math>200+240=440</math> mL of orange juice and <math>600+600=1200</math> mL of fluids, giving a fraction of <math>\frac{440}{1200} = \boxed{\textbf{(C)}\ \frac{11}{30}}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/Dit3DvLEVJY Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2004|num-b=15|num-a=17}} | {{AMC8 box|year=2004|num-b=15|num-a=17}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 00:50, 25 March 2022
Contents
Problem
Two mL pitchers contain orange juice. One pitcher is full and the other pitcher is full. Water is added to fill each pitcher completely, then both pitchers are poured into one large container. What fraction of the mixture in the large container is orange juice?
Solution
The first pitcher contains mL of orange juice. The second pitcher contains mL of orange juice. In the large pitcher, there is a total of mL of orange juice and mL of fluids, giving a fraction of .
Video Solution
https://youtu.be/Dit3DvLEVJY Soo, DRMS, NM
See Also
2004 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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